where $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ and $\rho = \sigma + \gamma i$ is a non-trivial zero of $\zeta(s)$ (i.e. $\gamma_n$ is the imaginary part of the n-th $\rho)$).

Riemann had already conjectured this in 1859 and since he needed an 'always real' function for his next thought steps, he assumed $s=\frac12 + t i$ and defined the function $\Xi(t)=\xi(s)$ that has been proven to be equal to:

$$\Xi(t)= \Xi(0)\prod_\gamma\left(1-\frac{t^2}{\gamma^2}\right)$$

It is known that $\xi(s)=\xi(1-s)$ and $\xi(s)=\overline{\xi(\overline{s})}$ and this implies that $\xi(s)\xi(\overline{s})$ must be real. Also known is that when there is a non-trivial zero $\rho$ lying off the critical line, then also $1-\rho, \overline{\rho}, \overline{1-\rho}$ must be zeros and their product will always be real:

Numerical tests indicate that the conjecture is correct, but I am obvioulsy keen to find a proof (I do realise this is a way too big of a question to ask here!). Does anybody have a link that explains how $\Xi(t)= \Xi(0)\prod_\gamma\left(1-\frac{t^2}{\gamma^2}\right)$ has been derived? Also appreciate any other thoughts/steers on how to further progress this conjecture.

This seems to be fit way more for overflow than for SE, though I'm happy to see it here.
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DonAntonioJan 6 '13 at 15:52

Mangoldt has $\xi(t)= \xi(0)\cdot \prod(1 -t^2/\alpha_{\nu^2}).$ Are you saying $\Xi$ is some other function?
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danielJan 6 '13 at 16:25

Your $\gamma$ is Mangoldt's $\alpha.$ He cites Hadamard for this relation so it must be in one of Hadamard's better-known papers?
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danielJan 6 '13 at 16:41

Indeed Daniel. I guess that's the formula I am after. Keen to find the proof of it in Hadamard's papers! P.S. There is quite a bit of confusion on the use upper and lower case xi. Riemann confusingly used $\xi(t)$ for $\Xi(t)$ where the latter has become the convention.
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AgnoJan 6 '13 at 16:45

2 Answers
2

Hoping to redeem my previous incorrect answer, I offer the following. In an article about Riemann's paper in which he discusses the problem of Riemann mentioned in Edwards' book, Fresan gives your relation on page 26 and says:

Daniel. I have Edwards' book on my lap and open at page 20. It only shows the product formula for $\xi(s)$ i.e. the lower case version. I am keen to find out how the the product for the $\Xi(t)$, i.e. upper case version assuming $s=\frac12 +t i$, has been derived.
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AgnoJan 6 '13 at 16:10

Merci bien, Daniel :-) Dusting off my French. You are right. P26 seems to outline the trick for $s=\frac12+ti$! Now need to check whether the same logic works for $P(a+ti)$ that is equivalent to my $\xi(a+s i)$.
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AgnoJan 6 '13 at 18:38

@user46149: Appreciate the up-vote. You should probably wait a bit before accepting because someone may provide a more authoritative answer. I am not an expert but there are many on this site.
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danielJan 6 '13 at 19:05

Daniel. Yes! This indeed does the trick. Many thanks. I can now prove that when I f.i. assume that all zeros in the infinite product of $\left(\frac{1}{\rho}\right)\left(\frac{1}{\overline{\rho}}\right)$ lie on the line $\rho=\frac13$, that the formula for $\Xi(t)_\frac13$ still perfectly holds :-) Landau's Handbuch is also correct that $\Xi_\frac12(z) = \Xi_\frac12(-z)$ but this is not true for other values of $a$ !
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AgnoJan 6 '13 at 19:09

For completeness' sake and to avoid confusion, I can now confirm that the conjecture (even though the numerical results are very close) must be false. It is easy to see that when:

$$\left(1-\frac{t^2}{\gamma^2}\right)$$

would be independent of $a$, then it could be expressed as:

$$\frac{\Xi(t)}{\Xi(0)}$$

This function has the same zeros as $\zeta(s)$ and therefore its square can not be equal to $\dfrac{\Xi_a(t)\Xi_a(-t)}{\xi(a)^2}$ since the latter doesn't have zeros at those spots (otherwise the RH would also have been immediately proven false).